CSIR-UGC National Eligibility Test (NET) for Junior Research Fellowship and Lecturer-ship
PREVIOUS SOLVED QUESTIONS – 2012 June CSIR – NET EXAM
This time CSIR does not allow candidates to carry questions with them. We have collected maximum questions from our candidates (memory based).
1. Ice skates can easily slopes down in ice. Why ?
a) Temperature increases b. Ice melts under the skates c)…………………………….
Expln:- Pressure depends on two things: force and area
Pressure = Force /Area
Notice that a large force might only create a small pressure if it’s spread out over a wide area. Also, a small force can create a big pressure if the area is tiny. When the area is small, a moderate force can create a very large pressure. This is why a sharp knife is good at cutting things.
Ice skates have sharp edges, and thus a small area in contact with the ice This means that your weight creates a very large pressure on the ice, far more if you were standing in ordinary shoes.
Ice has an unusual property: that is, it can melt under pressure, even if it’s below OoC. When you are ice skating, you are actually skating on a layer of water that have just melted. This layer of water quickly re-freezes when you move on.
2. A mendelian cross involves two characters, Tall-Dwarf and Red-White. F1 progeny produced tall red and dwarf white plants in equal ratio. Then what is the genotype of the parents ?
a) TtRr x ttrr b) TTRR X ttrr c) Tt.r x ttrr d) TtRr x TtRr
Expln:- Whenever the offspring obtained is in equal ratio, one of the parents always be heterozygous (TtRr) and the other will be recessive (homozygous). This is an example of test cross.
3. If half moon is in the waning phase. Then it is arising from
a. western horizon b. eastern horizon c.45 from east d.45 degree from west
4. The ends of rope are fixed to two pegs such that orpe remains slack. A pencile is placed against rope and moved, such that the rope always remains taut. the shape of the curve traced by pencile would be a part of
a. a circle b. an ellipse c.a square d.a triangle
5. The angles of a right angled triangle shaped garden are in arithmetic progression and the smallest side is 10m. Then the total length of the fencing of the garden in meter is
a. 60m b. 47.32m c.12.68m d.22.68m
Expln:- Sin 30° = opposit die / Hypotenuse =10/AC
∴ 1/2 = 10/AC ie; AC = 10 x 2 = 20
Tan 60° = opp.side/ adjacent side = BC / 10
Tan 60° = √3 = 1.732
∴ BC = 10 x 1.732 = 17.32
∴ The total length of the fencing of the garden is 10+20+17.32= 47.32m
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